Problem: $g(t) = -2t^{2}-4t+1+f(t)$ $h(x) = 6x^{2}+2x-f(x)$ $f(x) = -7x^{3}+7x^{2}+2$ $ g(h(0)) = {?} $
Solution: First, let's solve for the value of the inner function, $h(0)$ . Then we'll know what to plug into the outer function. $h(0) = 6(0^{2})+(2)(0)-f(0)$ To solve for the value of $h$ , we need to solve for the value of $f(0)$ $f(0) = -7(0^{3})+7(0^{2})+2$ $f(0) = 2$ That means $h(0) = 6(0^{2})+(2)(0)-2$ $h(0) = -2$ Now we know that $h(0) = -2$ . Let's solve for $g(h(0))$ , which is $g(-2)$ $g(-2) = -2(-2)^{2}+(-4)(-2)+1+f(-2)$ To solve for the value of $g$ , we need to solve for the value of $f(-2)$ $f(-2) = -7(-2)^{3}+7(-2)^{2}+2$ $f(-2) = 86$ That means $g(-2) = -2(-2)^{2}+(-4)(-2)+1+86$ $g(-2) = 87$